# Amenability by Alan L. T. Paterson PDF

By Alan L. T. Paterson

ISBN-10: 0821809857

ISBN-13: 9780821809853

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Sample text

Remark 1. Notice that the formula can also be written · V) − ( ×( × V) = 2 V. ♦ Remark 2. We note for completeness ∂Vx ∂Vy ∂Vz + + ∂x ∂y ∂z 2 2 2 ∂ 2 Vx ∂ Vx ∂ Vz ∂ 2 Vy ∂ Vy ∂ 2 Vz + + + + + e x ∂x2 ∂x∂y ∂x∂z ∂x∂y ∂y 2 ∂y∂z 2 2 2 ∂ Vx ∂ Vy ∂ Vz + + + ♦ ez . 2 Find div V and rot V for each of the following vector ﬁelds on R3 . 1) V(x, y, z) = (xz, −y 2 , 2x2 y). 2) V(x, y, z) = (z + sin y, −z + cos y, 0). 3) V(x, y, z) = (exy , cos(xy), cos(xz 2 )). 4) V(x, y, z) = (x2 + yz, y 2 + xz, z 2 + xy).

6 The parametric description of F1 is chosen as u 2 + v 2 ≤ a2 , a2 − u2 − v 2 , r(u, v) = u, v, thus ∂r = ∂u 1, 0, − √ a2 u − u2 − v 2 ∂r = ∂v and 0, 1, − √ a2 v − u2 − v 2 , from which we get the normal vector ex ey ez 1 0 u −√ 2 a − u2 − v 2 0 1 v −√ 2 a − u2 − v 2 N(u, v) = 1 =√ 2 a − u2 − v 2 a2 − u2 − v 2 , u, v, which is clearly pointing away from the body, because the Z-coordinate is +1. If we put B = (u, v) | u2 + v 2 < a2 , it follows from (x, y, z) = u, v, ﬂux(F1 ) = F1 V · n dS = (2u+2v, 2v+ = B ·√ = B B 1 B 1 √ {2u2 +2uv+2v 2 +v 2 a −u2 −v 2 0 2a2 √ − a2 − t √ 2 = π 4a2 a2 − a3 3 a2 = a2 −u2 −v 2 a2 − u2 − v 2 }dudv a2 + u2 + v 2 √ du dv + 0 = a2 − u 2 − v 2 a2 =π a2 −u2 −v 2 +2u) a2 −u2 −v 2 )dudv (u, v, +(a2 − u2 − v 2 ) + 2u = a2 − u2 − v 2 that V(u, v) · N(u, v) du dv a2 −u2 −v 2 , a2 −u2 −v 2 √ 2π 0 a a2 + a2 − 2 0 − t dt = π −4a 2 2 a2 · d 2 −t+ ( 3 a2 dϕ = π a2 − t) 3 0 a2 +t √ dt a2 −t a2 0 10π 3 a .

5 –1 Figure 18: The cut of the meridian half plane for a = 1. I By using semi polar coordinates we obtain that az ≥ 2 − a2 og z 2 + 2 ≤ a2 , and the meridian half plane becomes like shown on the ﬁgure. com 64 Calculus 2c-9 Gauβ’s theorem and div V = 2 + 2 + 1 = 5, it follows from Gauß’s theorem that ﬂux(F) = ﬂux(F1 ) + ﬂux(F2 ) = F V · n dS = div V dΩ = 5 vol(Ω) = Ω 35π 3 a . 6 The parametric description of F1 is chosen as u 2 + v 2 ≤ a2 , a2 − u2 − v 2 , r(u, v) = u, v, thus ∂r = ∂u 1, 0, − √ a2 u − u2 − v 2 ∂r = ∂v and 0, 1, − √ a2 v − u2 − v 2 , from which we get the normal vector ex ey ez 1 0 u −√ 2 a − u2 − v 2 0 1 v −√ 2 a − u2 − v 2 N(u, v) = 1 =√ 2 a − u2 − v 2 a2 − u2 − v 2 , u, v, which is clearly pointing away from the body, because the Z-coordinate is +1.